# A Inquiry regarding a matrix tensor notation...

#### berlinspeed

Summary
When comparing $g_{\mu\nu}$ with $\eta_{\mu\nu}$ in curved spacetime.
So if $P_{0}$ is an event, and I have $\mathcal {g_{\mu\nu}(P_{0})}=0$ and $\mathcal {g_{\mu\nu,\alpha\beta}(P_{0})}\neq0$, does this notation mean $\partial\alpha\partial\beta$ or simply $\partial(\alpha\beta)$? And what is the significance of it? Why can't it be zero in curved spacetime?

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#### Cryo

Gold Member
How about defining your terms? All of them? Also if you want to talk about curved space-time, please specify which derivatives you are talking about. Simple partial derivatives, covariant derivatives, if covariant, are you using Levi-Civita connection. etc

#### PeterDonis

Mentor
if $P_{0}$ is an event, and I have $\mathcal {g_{\mu\nu}(P_{0})}=0$
This makes no sense; the metric never vanishes. I think what you meant to write here is $\mathcal {g_{\mu\nu}(P_{0})}=\eta_{\mu \nu}$.

does this notation mean $\partial\alpha\partial\beta$ or simply $\partial(\alpha\beta)$?
The notation $\mathcal {g_{\mu\nu, \alpha \beta}(P_{0})}$ means $\mathcal {\partial_\alpha \partial_\beta g_{\mu\nu}(P_{0})}$. I have no idea what you mean by $\partial\alpha\partial\beta$ or $\partial(\alpha\beta)$.

#### PeterDonis

Mentor
what is the significance of it?
What is the significance of what?

It would help if you would give a specific reference for where you are getting this from. Also it would help to have some idea of how much background you have in GR; you marked this thread as "A" level but the questions you are asking don't indicate that you have that level of background knowledge.

#### Orodruin

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You can always find a coordinate system such that $g_{\mu\nu} = \eta_{\mu\nu}$ and the Christoffel symbols vanish at a given point $p$. However, in such a coordinate system, the Riemann curvature tensor at $p$ takes the form
$$R^d_{cab} = \partial_a \Gamma^d_{bc} - \partial_b \Gamma^d_{ac}.$$
Inserting the explicit form of the Christoffel symbols and lowering the $d$-index leads to
$$R_{abcd} = \frac{1}{2}(g_{bc,ad} + g_{ad,bc} - g_{bd,ac} - g_{ac,bd}).$$
Therefore, if $g_{\mu\nu,\alpha\beta} = 0$, then the curvature tensor would be identically equal to zero at $p$, i.e., the manifold would be flat at $p$, contrary to the assertion that the manifold is curved at $p$.

#### PeterDonis

Mentor
You can always find a coordinate system such that $g_{\mu\nu}$ and the Christoffel symbols vanish at a given point $p$.
Such that $g_{\mu \nu} = \eta_{\mu \nu}$ and the Christoffel symbols vanish at a given point $p$. The metric $g_{\mu \nu}$ does not vanish.

#### Orodruin

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Such that $g_{\mu \nu} = \eta_{\mu \nu}$ and the Christoffel symbols vanish at a given point $p$. The metric $g_{\mu \nu}$ does not vanish.
Meh, that is what happens when you edit your post too much before posting, you remove things that were needed to make it correct ...

Edit: Fixed ...

#### berlinspeed

You can always find a coordinate system such that $g_{\mu\nu} = \eta_{\mu\nu}$ and the Christoffel symbols vanish at a given point $p$. However, in such a coordinate system, the Riemann curvature tensor at $p$ takes the form
$$R^d_{cab} = \partial_a \Gamma^d_{bc} - \partial_b \Gamma^d_{ac}.$$
Inserting the explicit form of the Christoffel symbols and lowering the $d$-index leads to
$$R_{abcd} = \frac{1}{2}(g_{bc,ad} + g_{ad,bc} - g_{bd,ac} - g_{ac,bd}).$$
Therefore, if $g_{\mu\nu,\alpha\beta} = 0$, then the curvature tensor would be identically equal to zero at $p$, i.e., the manifold would be flat at $p$, contrary to the assertion that the manifold is curved at $p$.
Thanks for the explanation however my question was simply regarding the notation of $\mathcal {g_{\mu\nu,\alpha\beta}(P_{0})}\neq0$ -- does the $,\alpha\beta$ signify $\partial_{\alpha}\partial_{\beta}$ or $\partial_{(\alpha\beta)}$, now it seems to be the former..

Last edited:

#### Orodruin

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Thanks for the explanation however my question was simply regarding the notation of $\mathcal {g_{\mu\nu,\alpha\beta}(P_{0})}\neq0$ -- does the $,\alpha\beta$ signify $\partial_{\alpha}\partial_{\beta}$ or $\partial_{(\alpha\beta)}$, now it seems to be the former..
Summary: When comparing $g_{\mu\nu}$ with $\eta_{\mu\nu}$ in curved spacetime.

Why can't it be zero in curved spacetime?

#### Ibix

Thanks for the explanation however my question was simply regarding the notation of $\mathcal {g_{\mu\nu,\alpha\beta}(P_{0})}\neq0$ -- does the $,\alpha\beta$ signify $\partial_{\alpha}\partial_{\beta}$ or $\partial_{(\alpha\beta)}$, now it seems to be the former..
As noted above, $g_{\mu\nu,\alpha\beta}$ is a shorthand notation for $\partial_\alpha\partial_\beta g_{\mu\nu}$, which in turn is shorthand for $$\frac{\partial}{\partial x^\alpha}\frac{\partial}{\partial x^\beta}g_{\mu\nu}$$You seem to me to be trying to ask if it might instead mean$$\frac{\partial^2}{\partial x^\alpha\partial x^\beta}g_{\mu\nu}$$But that's exactly the same thing.

#### berlinspeed

As noted above, $g_{\mu\nu,\alpha\beta}$ is a shorthand notation for $\partial_\alpha\partial_\beta g_{\mu\nu}$, which in turn is shorthand for $$\frac{\partial}{\partial x^\alpha}\frac{\partial}{\partial x^\beta}g_{\mu\nu}$$You seem to me to be trying to ask if it might instead mean$$\frac{\partial^2}{\partial x^\alpha\partial x^\beta}g_{\mu\nu}$$But that's exactly the same thing.
No I thought it was the product of $\alpha$ and $\beta$ but it's not, which wouldn't make much sense.

"Inquiry regarding a matrix tensor notation..."

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